Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 1}{x + 1} = \dfrac{x + 3}{x + 1}$
Answer: Multiply both sides by $x + 1$ $ \dfrac{x^2 + 1}{x + 1} (x + 1) = \dfrac{x + 3}{x + 1} (x + 1)$ $ x^2 + 1 = x + 3$ Subtract $x + 3$ from both sides: $ x^2 + 1 - (x + 3) = x + 3 - (x + 3)$ $ x^2 + 1 - x - 3 = 0$ $ x^2 - 2 - x = 0$ Factor the expression: $ (x - 2)(x + 1) = 0$ Therefore $x = 2$ or $x = -1$ At $x = -1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -1$, it is an extraneous solution.